Find the x and y Intercepts f(x)=-(1/3)^(x+4)

$f (x) = - {(\frac{1}{3})}^{x + 4}$

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = - {(\frac{1}{3})}^{x + 4}$

Solve the equation.

Rewrite the equation as $- {(\frac{1}{3})}^{x + 4} = 0$ .

$- {(\frac{1}{3})}^{x + 4} = 0$

Simplify $- {(\frac{1}{3})}^{x + 4}$ .

Apply the product rule to $\frac{1}{3}$ .

$- \frac{1^{x + 4}}{3^{x + 4}} = 0$

One to any power is one.

$- \frac{1}{3^{x + 4}} = 0$

Set the numerator equal to zero.

$1 = 0$

Since $1 \neq 0$ , there are no solutions.

No solution

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

x-intercept(s): $None$

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = - {(\frac{1}{3})}^{(0) + 4}$

Solve the equation.

Remove parentheses.

$y = - {(\frac{1}{3})}^{0 + 4}$

Remove parentheses.

$y = - {(\frac{1}{3})}^{(0) + 4}$

Simplify $- {(\frac{1}{3})}^{(0) + 4}$ .

Add $0$ and $4$ .

$y = - {(\frac{1}{3})}^{4}$

Apply the product rule to $\frac{1}{3}$ .

$y = - \frac{1^{4}}{3^{4}}$

One to any power is one.

$y = - \frac{1}{3^{4}}$

Raise $3$ to the power of $4$ .

$y = - \frac{1}{81}$

y-intercept(s) in point form.

y-intercept(s): $(0, - \frac{1}{81})$

List the intersections.

x-intercept(s): $None$

y-intercept(s): $(0, - \frac{1}{81})$